The full braid group on $n$ strands $B_n$ admits a surjective homomorphism $p\colon\thinspace B_n\to \Sigma_n$ onto the symmetric group on $n$ letters, which takes a braid to the induced permutation of its ends. The kernel $P_n$ is well understood; it is the pure braid group on $n$ strands.

What about $p^{-1}(A_n)$, where $A_n$ is the alternating group on $n$ letters? Let me call this group $E_n$ for now, because I think it should be called the even braid group. However an internet search using this name (and others such as "orientation preserving braids", "positive braids" and so on) came up blank.

Does this group $E_n$ have a name, and has it been studied anywhere in the literature?

I would be particularly interested in computations of the cohomology rings of these groups.

Update:
It occurred to me that there was one obvious name I hadn't searched for, which was "alternating subgroups of braid groups". This led me to the following preprint,

which has a section on finding presentations for these groups (the alternating subgroup of the braid group associated to a Coxeter system $(G,S)$ is denoted $\mathcal{B}^+(G)$ in Section 5). So it seems that they do indeed appear in the literature, although not until surprisingly recently!

You have a lot of faith in me HW...thankfully I do know something about their cohomology rings...
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Andy PutmanFeb 7 '13 at 13:57

1

@Vladimir: I like to try to compute the topological complexity (in the sense of Farber arxiv.org/abs/math/0111197) of $K(\pi,1)$'s. My current idea is to use the trace and knowledge of the cohomology of $P_n$ to get information for $B_n$. My methods weren't working, but showed signs that they might if I had half as many elements...which led me to these $E_n$'s!
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Mark GrantFeb 7 '13 at 14:19

3 Answers
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I don't know if these groups have been studied before, but I can say something about their cohomology rings, at least over $\mathbb{Q}$. Namely, we have $H^k(E_n;\mathbb{Q}) = \mathbb{Q}$ if $k=0,1$ and $H^k(E_n;\mathbb{Q}) = 0$ for $k \geq 2$. Of course, this is the same as the cohomology of the ordinary braid group as computed by Arnold in

Recall that if $H$ is a finite-index normal subgroup of $G$, then $G$ acts on $H^k(H;\mathbb{Q})$ and using the transfer map we have that $H^k(G;\mathbb{Q})$ is equal to the invariants of this action.

For braid groups, the action of $B_n$ on $H^k(PB_n;\mathbb{Q})$ factors through an action of the symmetric group $S_n$, so $H^k(PB_n;\mathbb{Q})$ is a representation of $S_n$ and $H^k(B_n;\mathbb{Q})$ is the trivial subrepresentation $\{\text{$v \in H^k(PB_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in S_n$}\}$.

Let's now consider $E_n$. In this case, the above argument shows that $H^k(E_n;\mathbb{Q})$ is the subrepresentation $\{\text{$v \in H^k(PB_n;\mathbb{Q})$ $|$ $\sigma v = v$ for all $\sigma \in A_n$}\}$.

Now, representations of finite groups over $\mathbb{Q}$ decompose into direct sums of irreducible representations. The only two irreducible representations of $S_n$ that restrict to the identity on $A_n$ are the trivial representation and the alternating representation. As we said above, the trivial representation corresponds to $H^k(B_n;\mathbb{Q})$, so we conclude that
$$H^k(E_n;\mathbb{Q}) = W \oplus H^k(B_n;\mathbb{Q}),$$
where $W \subset H^k(PB_n;\mathbb{Q})$ is the direct sum of all alternating subrepresentations.

The above calculation is thus equivalent to the assertion that the alternating representation does not occur in $H^k(PB_n;\mathbb{Q})$. This follows from the calculation of $H^k(PB_n;\mathbb{Q})$ as a representation of $S_n$ which was done by in the paper "Coxeter group actions on the complement of hyperplanes and special involutions" by Felder-Velesov; see here.

The above ref to Felder-Velesov was suggested by Vladimir Dotsenko; I originally included the argument below, which only works for $n \gg k$.

It's quite hard to decompose $H^k(PB_n;\mathbb{Q})$ into irreducibles; however the paper "Representation Theory and Homological Stability" (see here) by Church and Farb introduces a recipe that they call "representation stability" which describes how the decomposition of $H^k(PB_{n+1};\mathbb{Q})$ into irreducibles can be constructed from the decomposition of $H^k(PB_n;\mathbb{Q})$ into irreducibles, at least for $n$ large. Their results are hard to summarize briefly, but they do imply that the alternating representation does not occur (it is not "stable" in their sense), again at least for $n$ large.

That's very nice. A quick remark on what you say ("it's quite hard to decompose $H^k(PB_n;\mathbb{Q})$ into irreducibles"): a description of $H^*(PB_n;\mathbb{Q})$ as a representation is done by Felder and Veselov (e.g. arxiv.org/abs/math/0311190), they prove that it is isomorphic to $2 Ind^{S_n}_{\langle (12)\rangle}(1)$. This clearly shows that the alternating representation does not occur at all (Frobenius reciprocity).
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Vladimir DotsenkoFeb 7 '13 at 16:23

@Vladimir Dotsenko : Thanks for letting me know about that paper; I had not seen it before! It shows that the "large $n$" condition I included is not necessary.
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Andy PutmanFeb 8 '13 at 21:29

As well as Andy's answer above, they have another property in common with the full braid groups $B_n$: they satisfy homological stability. There are two ways (that I know of) that one can prove this. The first is in the paper:

A sketch of the argument is as follows. It's enough to consider rational coefficients and $\mathbb{F}_p$ coefficients, for each prime $p$, separately. With $\mathbb{F}_2$ coefficients $E_n$ automatically inherits homological stability from $B_n$; this is actually true for any family of index-2 subgroups, or more generally double covering spaces, by using the mod-2 Gysin sequence and considering the double covers as 0-sphere bundles. (This only works for mod-2 coefficients since in general there is no Gysin sequence for 0-sphere bundles.)

When $F$ is a field of characteristic not 2, the homology $H_k(E_n;F)$ splits as
$$
H_k(E_n;F) \cong H_k(B_n;F) \oplus H_k(B_n;F^{(-1)})
$$
where $F^{(-1)}$ is the local coefficient system where the odd braids act by multiplication by $-1$. A model of the classifying space of $B_n$ is the configuration space $C_n(\mathbb{R}^2)$ of $n$ unordered points on the plane. There is a result of Boedigheimer, Cohen, Milgram and Taylor which calculates $H_k(C_n(M);F^{(-1)})$ in the range $k<\mathrm{dim}(M)n$ for any even-dimensional manifold $M$, and the above paper uses this to compute the $H_k(B_n;F^{(-1)})$ summand in this range. It turns out to be zero in a stable range, and so homological stability for $E_n$ follows from homological stability for $B_n$.

The calculations of the Guest-Kozlowsky-Yamaguchi paper actually work more generally, for configuration spaces on any open connected surface, so homological stability is also true for ``alternating surface braid groups''.

The stable range is worse than that for the full braid groups, however: $H_k(B_n;\mathbb{Z})$ is independent of $n$ for approximately $n\geq 2k$, whereas $H_k(E_n;\mathbb{Z})$ only becomes independent of $n$ for approximately $n\geq 3k$. The obstruction to $E_n$ having the better range lies entirely in the 3-torsion.

The second way I know of proving homological stability for the alternating braid groups is kind of a shameless plug, as it's something that I wrote (apologies if this is inappropriate; this is my first MO answer). It works more generally for what could be called ``alternating configuration spaces'' (but which are actually called oriented configuration spaces since that's what they were called by the GKY paper above). It's in the preprint

Just for historical completeness, I should also mention that homological stability for the alternating groups $A_n$ (which can be thought of as the alternating braid groups on $\mathbb{R}^\infty$ if one is so inclined) was proved much earlier, as Proposition A on page 130 of the paper:

I don't think it's the same, at least superficially. In the OP's question, braids do not necessarily are on the even number of strands.
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Vladimir DotsenkoFeb 7 '13 at 13:56

2

I think Vladimir is right, the thesis you link seems to be using "even" only in the sense of number of strands. In particular I couldn't see anything about the alternating group in there.
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Mark GrantFeb 7 '13 at 14:14